ÌâÄ¿ÄÚÈÝ

5£®£¨1£©ÊµÑé²âµÃ16g¼×´¼[CH3OH£¨l£©]ÔÚÑõÆøÖгä·ÖȼÉÕÉú³É¶þÑõ»¯Ì¼ÆøÌåºÍҺ̬ˮʱÊͷųö363.25kJµÄÈÈÁ¿£¬ÊÔд³ö¼×´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ£º
CH3OH £¨l£©+$\frac{3}{2}$ O2£¨g£©=CO2£¨g£©+2H2O£¨l£©¡÷H=-726.5KJ/mol£®
£¨2£©´Ó»¯Ñ§¼üµÄ½Ç¶È·ÖÎö£¬»¯Ñ§·´Ó¦µÄ¹ý³Ì¾ÍÊÇ·´Ó¦ÎïµÄ»¯Ñ§¼ü±»ÆÆ»µºÍÉú³ÉÎïµÄ»¯Ñ§¼üµÄÐγɹý³Ì£®ÒÑÖª·´Ó¦£ºN2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H=a kJ•mol-1£®ÓйؼüÄÜÊý¾ÝÈç±í£º
»¯Ñ§¼üH-HN-HN¡ÔN
¼üÄÜ£¨kJ•mol-1£©436391945
ÊÔ¸ù¾Ý±íÖÐËùÁмüÄÜÊý¾Ý¹ÀËãaµÄÊýÖµ-93£®
£¨3£©ÒÀ¾Ý¸Ç˹¶¨ÂÉ¿ÉÒÔ¶ÔijЩÄÑÒÔͨ¹ýʵÑéÖ±½Ó²â¶¨µÄ»¯Ñ§·´Ó¦µÄ·´Ó¦ÈȽøÐÐÍÆË㣮ÒÑÖª£º
C£¨s£¬Ê¯Ä«£©+O2£¨g£©¨TCO2£¨g£©¡÷H1=-393.5kJ•mol-1
2H2£¨g£©+O2£¨g£©¨T2H2O£¨l£©¡÷H2=-571.6kJ•mol-1
2C2H2£¨g£©+5O2£¨g£©¨T4CO2£¨g£©+2H2O£¨l£©¡÷H3=-2599kJ•mol-1
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¼ÆËã298KʱÓÉC£¨s£¬Ê¯Ä«£©ºÍH2£¨g£©Éú³É1mol C2H2£¨g£©·´Ó¦µÄ·´Ó¦ÈÈΪ£º
¡÷H=+226.7 KJ/mol£®
£¨4£©ÔÚ΢ÉúÎï×÷ÓõÄÌõ¼þÏ£¬NH4+¾­¹ýÁ½²½·´Ó¦±»Ñõ»¯³ÉNO3-£®Á½²½·´Ó¦µÄÄÜÁ¿±ä»¯Ê¾
ÒâͼÈçͼ£º

µÚÒ»²½·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¨Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±£©£¬Ô­ÒòÊÇ·´Ó¦Îï×ÜÄÜÁ¿¸ßÓÚÉú³ÉÎï×ÜÄÜÁ¿£®

·ÖÎö £¨1£©È¼ÉÕÈÈÊÇ1mol¿ÉȼÎïÍêȫȼÉÕÉú³ÉζÈÑõ»¯Îï·Å³öµÄÈÈÁ¿£¬¸ù¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨¿ÉÖª£¬»¯Ñ§¼ÆÁ¿ÊýÓë·´Ó¦ÈȳÉÕý±È£¬²¢×¢Òâ±êÃ÷ÎïÖʵľۼ¯×´Ì¬À´½â´ð£»
£¨2£©·´Ó¦ÈȵÈÓÚ·´Ó¦ÎïµÄ×ܼüÄÜ-Éú³ÉÎïµÄ×ܼüÄÜÇóË㣻
£¨3£©¿ÉÒÔÏȸù¾Ý·´Ó¦ÎïºÍÉú³ÉÎïÊéд»¯Ñ§·½³Ìʽ£¬¸ù¾Ý¸Ç˹¶¨ÂɼÆËã·´Ó¦µÄìʱ䣬×îºó¸ù¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨À´ÊéдÈÈ»¯Ñ§·½³Ìʽ£»
£¨4£©µ±·´Ó¦ÎïµÄ×ÜÄÜÁ¿´óÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£¬·´Ó¦ÊÇ·ÅÈȵģ®

½â´ð ½â£º£¨1£©16gCH3OHÔÚÑõÆøÖÐȼÉÕÉú³ÉCO2ºÍҺ̬ˮ£¬·Å³ö363.25kJÈÈÁ¿£¬32g¼´1molCH3OHÔÚÑõÆøÖÐȼÉÕÉú³ÉCO2ºÍҺ̬ˮ£¬·Å³ö726.5kJÈÈÁ¿£¬
Ôò¡÷H=-726.5KJ/mol£¬ÔòȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCH3OH £¨l£©+$\frac{3}{2}$ O2£¨g£©=CO2£¨g£©+2H2O£¨l£©¡÷H=-726.5KJ/mol£¬
¹Ê´ð°¸Îª£ºCH3OH £¨l£©+$\frac{3}{2}$ O2£¨g£©=CO2£¨g£©+2H2O£¨l£©¡÷H=-726.5KJ/mol£»
£¨2£©N2£¨g£©+3H2£¨g£©???2NH3£¨g£©¡÷H=945kJ•mol-1+436kJ•mol-1¡Á3-391kJ•mol-1¡Á6=-93kJ•mol-1=a kJ•mol-1£¬Òò´Ëa=-93£¬
¹Ê´ð°¸Îª£º-93£»   
£¨3£©ÒÑÖª£º¢ÙC £¨s£¬Ê¯Ä«£©+O2£¨g£©=CO2£¨g£©¡÷H1=-393.5kJ•mol-1£»
¢Ú2H2£¨g£©+O2£¨g£©=2H2O £¨l£©¡÷H2=-571.6kJ•mol-1£»
¢Û2C2H2£¨g£©+5O2£¨g£©¨T4CO2£¨g£©+2H2O £¨l£©¡÷H2=-2599kJ•mol-1£»
2C £¨s£¬Ê¯Ä«£©+H2£¨g£©=C2H2£¨g£©µÄ·´Ó¦¿ÉÒÔ¸ù¾Ý¢Ù¡Á2+¢Ú¡Á$\frac{1}{2}$-¢Û¡Á$\frac{1}{2}$µÃµ½£¬
ËùÒÔ·´Ó¦ìʱä¡÷H=2¡Á£¨-393.5kJ•mol-1£©+£¨-571.6kJ•mol-1£©¡Á$\frac{1}{2}$-£¨-2599kJ•mol-1£©¡Á$\frac{1}{2}$=+226.7kJ•mol-1£¬
¹Ê´ð°¸Îª£º+226.7 KJ/mol£»
£¨4£©ÒòΪ¡÷H=-273kJ/mol£¼0£¬Ôò·´Ó¦Îª·ÅÈÈ·´Ó¦£¬·´Ó¦ÎïµÄ×ÜÄÜÁ¿´óÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£¬
¹Ê´ð°¸Îª£º·ÅÈÈ£¬·´Ó¦Îï×ÜÄÜÁ¿¸ßÓÚÉú³ÉÎï×ÜÄÜÁ¿£®

µãÆÀ ±¾Ì⿼²é·´Ó¦ÈȵļÆË㣬עÒâ¸ù¾ÝÒÑÖªÈÈ»¯Ñ§·½³ÌʽÀûÓøÇ˹¶¨ÂɼÆËã·´Ó¦Èȵķ½·¨£¬´ðÌâʱעÒâÌå»á£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
10£®ÈçͼËùʾ£¬ºá×ø±êΪÈÜÒºµÄpH£¬×Ý×ø±êΪZn2+»ò[Zn£¨OH£©4]2-µÄÎïÖʵÄÁ¿Å¨¶ÈµÄ¶ÔÊý£®

»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÍùZnCl2ÈÜÒºÖмÓÈë×ãÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº£¬·´Ó¦µÄÀë×Ó·½³Ìʽ¿É±íʾΪZn2++4OH-¨T[Zn£¨OH£©4]2-£®
£¨2£©´ÓͼÖÐÊý¾Ý¼ÆËã¿ÉµÃZn£¨OH£©2µÄÈܶȻýKsp=1.00¡Á10-17£®
£¨3£©Ä³·ÏÒºÖк¬Zn2+£¬ÎªÌáÈ¡Zn2+¿ÉÒÔ¿ØÖÆÈÜÒºÖÐpHµÄ·¶Î§ÊÇ8.0£¼pH£¼12.0£®
ij¹¤³§ÓÃÁùË®ºÏÂÈ»¯Ã¾ºÍ´Öʯ»ÒÖÆÈ¡µÄÇâÑõ»¯Ã¾º¬ÓÐÉÙÁ¿ÇâÑõ»¯ÌúÔÓÖÊ£¬Í¨¹ýÈçÏÂÁ÷³Ì½øÐÐÌá´¿¾«ÖÆ£®»ñµÃ×èȼ¼ÁÇâÑõ»¯Ã¾£®

£¨4£©²½Öè¢ÙÖмÓÈë±£ÏÕ·Û£¨Na2S2O4£©µÄ×÷Ó㺽«ÇâÑõ»¯Ìú»¹Ô­ÎªÇâÑõ»¯ÑÇÌú
£¨5£©ÒÑÖªEDTAÖ»ÄÜÓëÈÜÒºÖеÄFe2+·´Ó¦Éú³ÉÒ×ÈÜÓÚË®µÄÎïÖÊ£¬²»ÓëMg£¨OH£©2·´Ó¦£®ËäÈ»Fe£¨OH£©2ÄÑÈÜÓÚË®£¬µ«²½Öè¢ÚÖÐËæ×ÅEDTAµÄ¼ÓÈ룬×îÖÕÄܹ»½«Fe£¨OH£©2³ýÈ¥²¢»ñµÃ´¿¶È¸ßµÄMg£¨OH£©2£®Çë´Ó³ÁµíÈÜ½âÆ½ºâµÄ½Ç¶È¼ÓÒÔ½âÊÍFe£¨OH£©2Ðü×ÇÒºÖдæÔÚÈçÏÂÆ½ºâ£ºFe£¨OH£©2£¨s£©=Fe2+£¨aq£©+2OH-£¨aq£©µ±²»¶ÏµÎÈëEDTAʱ£¬EDTA½«½áºÏFe 2+´ÙʹƽºâÏòÓÒÒÆ¶¯¶øÊ¹Fe£¨OH£©2²»¶ÏÈܽ⣻£»
¢ô£®ÎªÑо¿²»Í¬·ÖÀëÌá´¿Ìõ¼þÏÂËùÖÆµÃ×èȼ¼ÁµÄ´¿¶È´Ó¶øÈ·¶¨×î¼ÑÌá´¿Ìõ¼þ£¬Ä³Ñо¿Ð¡×é¸÷È¡µÈÖÊÁ¿µÄÏÂÁÐ4×éÌõ¼þÏÂÖÆµÃµÄ×èȼ¼Á½øÐк¬ÌúÁ¿µÄ²â¶¨£¬½á¹ûÈç±í£º
¾«ÖÆ×èȼ¼ÁµÄÌõ¼þ×èȼ¼ÁÌúº¬Á¿
ÐòºÅÌá´¿ÌåϵζÈ/¡æ¼ÓÈëEDTAÖÊÁ¿/g¼ÓÈë±£ÏÕ·ÛÖÊÁ¿/gW£¨Fe£©/£¨10-4g£©
1400.050.057.63
2400.050.106.83
3600.050.106.83
4600.100.106.51
£¨6£©Èô²»¿¼ÂÇÆäËüÌõ¼þ£¬¸ù¾ÝÉϱíÊý¾Ý£¬ÖÆÈ¡¸ß´¿¶È×èȼ¼Á×î¼ÑÌõ¼þÊÇC£¨Ìî×Öĸ£©£®
¢Ù40¡æ¢Ú60¡æ¢ÛEDTAÖÊÁ¿ÎªO.05g   ¢ÜEDTAÖÊÁ¿Îª0.10g  ¢Ý±£ÏÕ·ÛÖÊÁ¿Îª0.05g¢Þ±£ÏÕ·ÛÖÊÁ¿Îª0.10g
A£®¢Ù¢Û¢ÝB£®¢Ú¢Ü¢ÞC£®¢Ù¢Ü¢ÞD£®¢Ú¢Û

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø